Still working on the puzzle titled A Cube of Resistors?

Imagine welding a cube of resistors, each resistor with the resistance of 1 Ohm. If you measured the resistance between opposite corners, what would it be?

I’ll give you a hint. Well, more of a hint, I’ll suggest some guidelines of how a clever method of how to solve it. Don’t read anymore if you want to figure it out on your own.

SPOILER ALERT

Do not read any further if you don’t want any clues on how to solve the A Cube of Resistors puzzle. Be an Electrical Engineer! Don’t cheat!

Alright, here is the hint. The cube has many planes of symmetry. Exploiting them provides a very elegant solution. I decided to redraw the color to illustrate how many resistors are in similar circumstances.

There is a symmetry of the puzzle. The different colored zones have similar properties.

For example, all the resistors corresponding to the blue lines have a common point, the red one, with the same potential. But, the other end of the blue lines must also have the same potential. Why? Just look further down… there is a symmetry to each line. Follow the path from each blue line to the other end, and you will they look the same. Thus, although technically the blue lines are not in parallel, the symmetry of the cube makes them be in the same potential on both ends so they can be treated as if they were in parallel.

A similar argument works for the 6 resistors corresponding to the black lines; they can all be thought as if they were in the same potential difference between their ends. Also, treat the 3 green lines in the same manner, as if they were parallel resistors.

Following this argument, the Cube of Resistors will have the same resistance as the following figure.

This is not a cube. Yet, it has the same resistance and potential changes as the Cube of Resistors.

This figure is obtained by collapsing all the points with the same potential in the cube to one. After all, if the have the same potential, you can short circuit them without changing anything, right?

Now, this problem is much easier to solve. Treat the blues as 3 resistors in parallel, followed in series by 6 in parallel, followed in series by 3 in parallel. Come on, you can do it!

If you want to check your final answer, read ahead. Otherwise, don’t as it will spoil the fun.

BIG SPOILER ALERT

Do not read any further if you don’t want the answer to A Cube of Resistors puzzle.

…

Correct answer to the puzzle coming ahead.

…

Don’t read any further if you don’t want to know the numerical answer.

5/6 of an Ohm

Where does the number come from? Well, the equivalent resistance of the 3 blue lines as if they were in parallel is 1/3 Ohm (three resistance in parallel can carry three times as much current as one of them on its own. By a similar argument, the equivalent resistance of the 6 black lines as if they were in parallel is 1/6 Ohm. The equivalent resistance of the 3 green lines as if they were in parallel is 1/3 Ohm.

Since each color is in series with each other, the resistances add.

1/3 Ohm + 1/6 Ohm + 1/3 Ohm = 5/6 Ohm.

—
“Let’s play this with balalaikas. Give me the biggest balaika! We were open about stuff, we could do that.”
-The Clash

Here is a puzzle for those of you with some knowledge of circuit analysis.

Imagine welding a cube of resistors, each resistor with the resistance of 1 Ohm. If you measured the resistance between opposite corners, what would it be?

Each black line represents a resistor of 1 Ohm. What is the equivalent resistance between the red points?

Hints:

Do it the hard way: Distort the cube into a two-dimensional surface and use the full machinery of equivalent resistance (Y-Delta transformations, etc) to solve all the loops.

Do it the fun way: Think of the particular geometry of the cube, and its symmetries. Think of the fundamentals of circuit analysis. You really don’t need to calculate much if you know how to visualize the problem.

—

“The art of making love, muffled up in furs, in the open air, with the thermometer at Zero, is a Yankee invention.”
-President John Quincy Adams

Previously, on Minus Two Fish:
P.A.M. Dirac was presented with a puzzle.

Three fishermen come back from the sea. Each collapses in their respective tent. Fisherman #1 wakes up and decides to get his share of the catch. He counts the fish, realizes it is not a number divisible by three, throws away one fish to the sea correcting the situation, and takes a third of the remaining fish into his tent. Fisherman #2 wakes up later and also decides he is going to get his share of the catch. Fisherman #2 wants a third of the fish he sees. It is not a multiple of three, but he throws away one fish and takes a third of the fish and goes to sleep. Fisherman #3 wakes up after, and does the same: he throws away one fish, takes a third of the fish, and goes into his tent.
What is the smallest number of fish for which this would happen?

The real solution was not revealed, until now.

[Dirac’s Fish Puzzle Solution Spoiler Warning]

The want to find what is the smallest number of fish at the beginning.

is the number of fish after Fisherman #1 throws away one. is the number of fish that Fisherman #1 takes for himself, leaving only . is the number of fish after Fisherman #2 throws away one. is the number of fish that Fisherman #2 leaves after he takes his share. is the number of fish after Fisherman #3 throws away one. is the number of fish that Fisherman #3 leaves after he takes his share.

The last number must be an integer , as it is implicit that only a whole number fish are left.
. Solving for ,
.

Now, must be a small integer. The easiest way to explain the solution is by trial and error. So, try , and you will see is not an integer. Try , … still is not an integer. Until you reach , where .
So, the smallest number of fish is 25.
Also, try in the equation to plug in , Dirac’s solution and you will see . That is why it is so weird, Dirac assumes that there are negative fish to start, and negative fish at the end.

This solution permits for infinite number of fishermen to come and throw away fish and take of the remaining fish as the number of fish at the end is the same as the beginning.

The nobel laurate in physics, extreme weirdo, inventor of the delta function () and my own personal hero Paul Adrien Maurice Dirac was presented with the following puzzle:

Three fishermen come back from the sea, celebrating the catch of the day.  They land their boat and set up camp. After much drinking (rum?), each collapses in their respective tent. Fisherman #1 wakes up and, after relieving himself, decides to get his share of the catch. He counts the fish, realizes it is not a number divisible by three, throws away one fish to the sea correcting the situation, and takes a third of the remaining fish into his tent. Fisherman #2 wakes up later, goes to pee too, and also decides he is going to get his share of the catch. Unaware that Fisherman #1 already took his part, Fisherman #2 wants a third of the fish he sees. It is not a multiple of three, but he throws away one fish and takes a third of the fish and goes to sleep. Fisherman #3 wakes up after, and does the same: he throws away one fish, takes a third of the fish, and goes into his tent.

What is the smallest number of fish for which this would happen?

I’m not going to spoil the puzzle by revealing the regular answer, but I can tell you Dirac’s answer.  A weird answer, but correct nevertheless.

Dirac prefered to sleep on his left side.

Dirac’s answer was minus two fish.